Partial Differential Equations with MATLAB

176 An Introduction to Partial Differential Equations with MATLAB©R

so the characteristics are the curves

y=2x^2 +c or 2x^2 −y=constant.

Now, let
ξ=x
η=2x^2 −y
or
x=ξ
y=2ξ^2 −η.

Then

ux=uξ+4xuη uy=−uη

and the transformed PDE is

uξ−u=0

with solution

u=g(η)eξ =g(2x^2 −y)ex

for arbitrary functiong. So the curves 2x^2 −y=care, indeed, characteristic.
More precisely, given the first-order linear PDE

a(x, y)ux+b(x, y)uy+c(x, y)u=f(x, y),

wedefinethe characteristic curves to be those curves satisfying

dy dx

=

b(x, y) a(x, y)

or, as is traditionally written,

dx a(x, y)

=

dy b(x, y)

. (5.6)

Definition 5.1Thecharacteristicsorcharacteristic curvesof the first-
order linear PDE

a(x, y)ux+b(x, y)uy+c(x, y)u=f(x, y)

are those curves satisfying the ODE

dx a(x, y)

=

dy b(x, y)

.